My first attempt would be to note that since $\arctan t$ is a strictly increasing function of $t$ we have that $\arctan a \gt \arctan b \implies a = \tan \arctan a \gt \tan \arctan b = b$.

Therefore we can write
$$\arctan (x+y) \gt y + \arctan x \iff x+y \gt \tan {(y + \arctan x)}$$
Then use $$\tan (A+B) = {\tan A + \tan B \over 1 - \tan A \tan B}$$